Storing database resource links

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  • Ney André de Mello Zunino
    #1

    Storing database resource links

    Hello.

    Could anyone explain to me why the following code fails?

    // =====Begin sample code=====

    global $blog;

    define("dbHost" , "localhost" );
    define("dbUser" , "ney");
    define("dbPassw ord", "xxx");
    define("dbDatab ase", "blog");

    function dbConnect()
    {
    // Stores the database resource link in the global array
    $blog["db"] = mysql_connect(c onstant("dbHost "),
    constant("dbUse r"),
    constant("dbPas sword"))
    or die("DB connection error: " . mysql_error());
    mysql_select_db ('blog')
    or die("Error selecting blog database: " . mysql_error());
    }

    function dbDisconnect()
    {
    // Closes the connection represented by the resource link stored
    // in the global array
    mysql_close($bl og["db"]);
    }

    dbConnect();
    ..
    ..
    ..
    dbDisconnect();

    // =====End sample code=====

    Upon runtime, PHP produces the following warning:

    Warning: mysql_close(): supplied argument is not a valid MySQL-Link
    resource in /home/ney/Blog/db.php on line 22

    I would appreciate if anyone could point out my mistake and kindly
    explain the situation.

    Thank you.

    --
    Ney André de Mello Zunino
    Tags: None
    • ZeldorBlat
      #2
      Re: Storing database resource links


      Ney André de Mello Zunino wrote:
      Hello.
      >
      Could anyone explain to me why the following code fails?
      >
      // =====Begin sample code=====
      >
      global $blog;
      >
      define("dbHost" , "localhost" );
      define("dbUser" , "ney");
      define("dbPassw ord", "xxx");
      define("dbDatab ase", "blog");
      >
      function dbConnect()
      {
      // Stores the database resource link in the global array
      $blog["db"] = mysql_connect(c onstant("dbHost "),
      constant("dbUse r"),
      constant("dbPas sword"))
      or die("DB connection error: " . mysql_error());
      mysql_select_db ('blog')
      or die("Error selecting blog database: " . mysql_error());
      }
      >
      function dbDisconnect()
      {
      // Closes the connection represented by the resource link stored
      // in the global array
      mysql_close($bl og["db"]);
      }
      >
      dbConnect();
      .
      .
      .
      dbDisconnect();
      >
      // =====End sample code=====
      >
      Upon runtime, PHP produces the following warning:
      >
      Warning: mysql_close(): supplied argument is not a valid MySQL-Link
      resource in /home/ney/Blog/db.php on line 22
      >
      I would appreciate if anyone could point out my mistake and kindly
      explain the situation.
      >
      Thank you.
      >
      --
      Ney André de Mello Zunino
      You need to put global $blog; inside your functions. This tells PHP to
      "bind" the variable $blog inside the function to the global one.
      Adding that should fix it...although personally I prefer not to use
      globals.

        Comment

        • Ney André de Mello Zunino
          #3
          Re: Storing database resource links

          ZeldorBlat wrote:
          You need to put global $blog; inside your functions. This tells PHP to
          "bind" the variable $blog inside the function to the global one.
          Adding that should fix it...although personally I prefer not to use
          globals.
          Thanks for your quick reply. You are right: globals are evil, at least
          most of the time. I reworked the code, which now looks like:

          // =====Begin sample code=====

          function dbConnect()
          {
          $conn = mysql_connect(c onstant("dbHost "),
          constant("dbUse r"),
          constant("dbPas sword"))
          or die("DB connection error: " . mysql_error());
          mysql_select_db ('blog')
          or die("Error selecting blog database: " . mysql_error());
          return $conn;
          }

          function dbDisconnect($c onn)
          {
          mysql_close($co nn);
          }

          $blog["dbconn"] = dbConnect();
          ..
          ..
          ..
          dbDisconnect($b log["dbconn"]);

          // =====End sample code=====

          I guess it has improved a bit, hasn't it?

          Cheers!

          --
          Ney André de Mello Zunino

            Comment

            • ZeldorBlat
              #4
              Re: Storing database resource links


              Ney André de Mello Zunino wrote:
              ZeldorBlat wrote:
              >
              You need to put global $blog; inside your functions. This tells PHP to
              "bind" the variable $blog inside the function to the global one.
              Adding that should fix it...although personally I prefer not to use
              globals.
              >
              Thanks for your quick reply. You are right: globals are evil, at least
              most of the time. I reworked the code, which now looks like:
              >
              // =====Begin sample code=====
              >
              function dbConnect()
              {
              $conn = mysql_connect(c onstant("dbHost "),
              constant("dbUse r"),
              constant("dbPas sword"))
              or die("DB connection error: " . mysql_error());
              mysql_select_db ('blog')
              or die("Error selecting blog database: " . mysql_error());
              return $conn;
              }
              >
              function dbDisconnect($c onn)
              {
              mysql_close($co nn);
              }
              >
              $blog["dbconn"] = dbConnect();
              .
              .
              .
              dbDisconnect($b log["dbconn"]);
              >
              // =====End sample code=====
              >
              I guess it has improved a bit, hasn't it?
              >
              Cheers!

              --
              Ney André de Mello Zunino
              Yes -- much better :)

                Comment

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